Introducción

En teoría de números, un número se considera malvado si hay un número par de 1 en su representación binaria. En el desafío de hoy, estarás identificando si un número dado es malo o no.

Reto

Su trabajo consiste en escribir un programa completo o una función que acepte un número entero no negativo como entrada y salida (o retorno) independientemente de si ese número es malo o no.

• Puede generar cualquier valor verdadero si el número es malo, y cualquier valor falso si el número no es malo.
• Puede ingresar y enviar en cualquier formato aceptable .
• Las lagunas estándar no están permitidas.
• La secuencia OEIS A001969 es la secuencia que contiene todos los números malvados.
• Aquí hay una lista de los primeros 10000 números malvados, para referencia (¡y más casos de prueba!)
• Esta pregunta es el código de golf , así que cuanto más corto, mejor.
• No se desanime por las respuestas extremadamente cortas en idiomas de golf. Te animo a enviar en cualquier idioma que te guste.

• Aquí hay algunos casos de prueba:

  3 => True
  11 => False
  777 => True
  43 => True
  55 => False
  666 => False
  

La tabla de posiciones

Al final de la página hay un fragmento de pila que contiene una tabla de clasificación para esta pregunta. (Gracias, @ MartinEnder)

Para asegurarse de que su respuesta se muestre, comience con un título, usando la siguiente plantilla de Markdown:

# Language Name, N bytes

¿Dónde Nestá el tamaño de su envío? Si mejora su puntaje, puede mantener los puntajes antiguos en el título, tachándolos. Por ejemplo:

# Ruby, <s>104</s> <s>101</s> 96 bytes

Si desea incluir varios números en su encabezado (por ejemplo, porque su puntaje es la suma de dos archivos o desea enumerar las penalizaciones de la bandera del intérprete por separado), asegúrese de que el puntaje real sea el último número en el encabezado:

# Perl, 43 + 2 (-p flag) = 45 bytes

También puede hacer que el nombre del idioma sea un enlace que luego aparecerá en el fragmento de la tabla de clasificación:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes

/* Configuration */

var QUESTION_ID = 169724; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 81420; // This should be the user ID of the challenge author.

/* App */

var answers = [],
  answers_hash, answer_ids, answer_page = 1,
  more_answers = true,
  comment_page;

function answersUrl(index) {
  return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}

function commentUrl(index, answers) {
  return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER;
}

function getAnswers() {
  jQuery.ajax({
    url: answersUrl(answer_page++),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function(data) {
      answers.push.apply(answers, data.items);
      answers_hash = [];
      answer_ids = [];
      data.items.forEach(function(a) {
        a.comments = [];
        var id = +a.share_link.match(/\d+/);
        answer_ids.push(id);
        answers_hash[id] = a;
      });
      if (!data.has_more) more_answers = false;
      comment_page = 1;
      getComments();
    }
  });
}

function getComments() {
  jQuery.ajax({
    url: commentUrl(comment_page++, answer_ids),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function(data) {
      data.items.forEach(function(c) {
        if (c.owner.user_id === OVERRIDE_USER)
          answers_hash[c.post_id].comments.push(c);
      });
      if (data.has_more) getComments();
      else if (more_answers) getAnswers();
      else process();
    }
  });
}

getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

function getAuthorName(a) {
  return a.owner.display_name;
}

function process() {
  var valid = [];

  answers.forEach(function(a) {
    var body = a.body;
    a.comments.forEach(function(c) {
      if (OVERRIDE_REG.test(c.body))
        body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
    });

    var match = body.match(SCORE_REG);
    if (match)
      valid.push({
        user: getAuthorName(a),
        size: +match[2],
        language: match[1],
        link: a.share_link,
      });
    //else console.log(body);
  });

  valid.sort(function(a, b) {
    var aB = a.size,
      bB = b.size;
    return aB - bB
  });

  var languages = {};
  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function(a) {
    if (a.size != lastSize)
      lastPlace = place;
    lastSize = a.size;
    ++place;

    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
      .replace("{{NAME}}", a.user)
      .replace("{{LANGUAGE}}", a.language)
      .replace("{{SIZE}}", a.size)
      .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
    jQuery("#answers").append(answer);

    var lang = a.language;
    lang = jQuery('<a>' + lang + '</a>').text();

    languages[lang] = languages[lang] || {
      lang: a.language,
      lang_raw: lang,
      user: a.user,
      size: a.size,
      link: a.link
    };
  });

  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);

  langs.sort(function(a, b) {
    if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
    if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) return -1;
    return 0;
  });

  for (var i = 0; i < langs.length; ++i) {
    var language = jQuery("#language-template").html();
    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
      .replace("{{NAME}}", lang.user)
      .replace("{{SIZE}}", lang.size)
      .replace("{{LINK}}", lang.link);
    language = jQuery(language);
    jQuery("#languages").append(language);
  }

}

body {
  text-align: left !important;
  display: block !important;
}

#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}

#language-list {
  padding: 10px;
  width: 500px;
  float: left;
}

table thead {
  font-weight: bold;
}

table td {
  padding: 5px;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/all.css?v=ffb5d0584c5f">
<div id="language-list">
  <h2>Shortest Solution by Language</h2>
  <table class="language-list">
    <thead>
      <tr>
        <td>Language</td>
        <td>User</td>
        <td>Score</td>
      </tr>
    </thead>
    <tbody id="languages">

    </tbody>
  </table>
</div>
<div id="answer-list">
  <h2>Leaderboard</h2>
  <table class="answer-list">
    <thead>
      <tr>
        <td></td>
        <td>Author</td>
        <td>Language</td>
        <td>Size</td>
      </tr>
    </thead>
    <tbody id="answers">

    </tbody>
  </table>
</div>
<table style="display: none">
  <tbody id="answer-template">
    <tr>
      <td>{{PLACE}}</td>
      <td>{{NAME}}</td>
      <td>{{LANGUAGE}}</td>
      <td>{{SIZE}}</td>
      <td><a href="{{LINK}}">Link</a></td>
    </tr>
  </tbody>
</table>
<table style="display: none">
  <tbody id="language-template">
    <tr>
      <td>{{LANGUAGE}}</td>
      <td>{{NAME}}</td>
      <td>{{SIZE}}</td>
      <td><a href="{{LINK}}">Link</a></td>
    </tr>
  </tbody>
</table>

EDITAR: Creo que esta pregunta no es un duplicado de esto , porque mientras que esa pregunta es contar el número de unos, esta pregunta es si el número es par. Aunque puede lograr esta pregunta simplemente contando los bits, también hay otros enfoques .

Conjunto Z80 (8 bits), 2 bytes

El siguiente código solo funciona con valores de hasta 255:

; Input is given in register A.
; P flag is set if A is evil.
B7     or A
C9     ret

Versión de 16 bits (funciona en todos los casos de prueba), 3 bytes

Esto funciona con valores de hasta 65535.

; Input is given in BC.
; Output is the same as above.
78     ld A,B
A9     xor C
C9     ret

Si te sientes aventurero, puedes reducir 1 byte almacenando la entrada AyC así

      ld BC, 777
C5    push BC
F1    pop AF

y luego corriendo

A9    xor C
C9    ret

Sin embargo, esto pone la carga sobre la persona que llama, por lo que puede ser que los dos bytes ( push BCy pop AF) también se cuenten.

JavaScript (ES6), 18 bytes

f=n=>n?!f(n&~-n):1

Pruébalo en línea!

Explicación

La lógica bit a bit es así:

• Para enteros, ~-nes equivalente a -(-n)-1, por lo que solo es otra forma de hacerlo n-1. En ese caso particular, podríamos haberlo usado n-1.

• n & (n-1)elimina el bit menos significativo establecido en 1 en n porque la disminución de n convierte todos los 0 finales en 1 y borra el 1 que sigue inmediatamente (por propagación de acarreo), mientras deja todo lo demás sin cambios.

  Ejemplo para n = 24 (11000 en binario):

    11000 (24)                  11000 (24)
  -     1                   AND 10111 (23)
  -------                   ---------
  = 10111 (23)              =   10000 (16)
     ^                           ^
     |                           |
     +--- this bit is cleared ---+
  

Por lo tanto, procesamos tantas llamadas recursivas como haya 1 en la representación binaria de n , invirtiendo el resultado cada vez con !. La última llamada siempre devuelve 1 .

Ejemplos:

f(24) = !f(16) = !!f(0) = !!1 = true
f(7) = !f(6) = !!f(4) = !!!f(0) = !!!1 = false

Python 2 , 25 bytes

lambda n:int(bin(n),13)%2

Pruébalo en línea!

bin(n)da un resultado como '0b10101'. Al leer esto como un entero de base 13, obtenemos

que reduce el módulo 2 a ≡ 1 ⋅ 1 5 + 1 ⋅ 1 4 + 0 ⋅ 1 3 + 1 ⋅ 1 2 + 0 ⋅ 1 1 + 1 ⋅

Entonces int(bin(n),13)%2es igual a 1 + (número de unidades en bin(n)) módulo 2.

Si nes malo, entonces el resultado es 1; de lo contrario es 0.

Aprendí este truco de Noodle9 .

Japt -h!, 5 4 3 bytes

¤å^

Intentalo

Explicación

¤       :Convert to base-2 string
 å^     :Cumulatively reduce by XORing
        :Implicitly output the last element negated

C # (compilador interactivo de Visual C #) , 43 38 bytes

Golfed Pruébelo en línea!

i=>Convert.ToString(i,2).Sum(c=>c)%2<1

Sin golf

i => Convert.ToString( i, 2 ).Sum( c => c ) % 2 < 1

Código completo con pruebas

Func<Int32, Boolean> f = i => Convert.ToString( i, 2 ).Sum( c => c ) % 2 < 1;

Int32[] testCases = { 3, 11, 777, 43, 55 };

foreach( Int32 testCase in testCases ) {
    Console.Write( $" Input: {testCase}\nOutput: {f(testCase)}" );
    Console.WriteLine("\n");
}

Console.ReadLine();

Lanzamientos

• v1.1 - -5 bytes- Reemplazado CountporSum
• v1.0 - 43 bytes- Solución inicial.

Notas

• Ninguna

Bash (sin utilidades externas), 56 44 bytes

while(($1));do set $(($1/2)) $(($2+$1%2));done;!(($2%2))

(($1))&&exec $0 $[$1/2] $[$2+$1%2];!(($2%2))

Esto supone que el número se encuentra en $1, después de haber sido pasado como el primer argumento de línea de comando. También supone que este es un script de shell (para que pueda hacerlo exec).

Recurre, de una manera, usando exec $0, hasta que el número (in $1) llegue a cero, dividiéndolo por dos en cada iteración. También suma (in $2) la cantidad de veces que obtenemos un número que es impar. Al final, el número original era "malvado" si la suma $2no es impar.

Invocaciones de ejemplo:

$ ./script 3 && echo evil
evil

$ ./script 11 && echo evil

$ ./script 777 && echo evil
evil

$ ./script 43 && echo evil
evil

$ ./script 55 && echo evil

Para 0:

$ ./script 0 && echo evil
./script: line 1: ((: %2: syntax error: operand expected (error token is "%2")
evil

Resultado correcto, con un poco de extra en el lateral.

R , 37 26 bytes

!sum(scan()%/%2^(0:31))%%2

Pruébalo en línea!

Una alternativa a la respuesta de Robert S. , evita la división de bits incorporada pero termina siendo menos golfosa y gracias a JayCe y digEmAll termina siendo un poco más golfista.

Solo funciona para enteros positivos menores que $2^{31}-1$.

05AB1E, 4 bytes

bSOÈ

Try it online or verify all test cases.

Explanation:

b       # Convert to binary string
        #  i.e. 777 → 1100001001
 S      # Change it to a list of 0s and 1s
        #  i.e. 1100001001 → ['1','1','0','0','0','0','1','0','0','1']
  O     # Take the sum
        #  i.e. ['1','1','0','0','0','0','1','0','0','1'] → 4
   È    # Check if it's even (1 as truthy, 0 as falsey)
        #  i.e. 4 → 1

Stax, 4 bytes

:1|e

Run and debug it

:1|e Full program, implicit input-evaluation
:1   Count set bits
  |e Check if even

R, 99 98 44 34 28 bytes

-1 thanks to Kevin Cruijssen! -54 thanks to ngm! -10 thanks to Giuseppe! -6 thanks to JayCe!

!sum(intToBits(scan())>0)%%2

Try it online!

Alternatively, using the binaryLogic package (39 bytes):

!sum(binaryLogic::as.binary(scan()))%%2

C (gcc), 36 bytes

c;f(n){for(c=0;n;c++)n&=n-1;n=~c&1;}

Try it online!

Method from K&R https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan

Must be compiled with optimization level 0

Wolfram Language (Mathematica), 24 22 bytes

• Saved two bytes thanks to Misha Lavrov.

2∣DigitCount[#,2,1]&

Try it online!

PHP, 37 36 bytes

<?=1&~substr_count(decbin($argn),1);

To run it:

echo '<input>' | php -nF <filename>

Or Try it online!

Prints 1 for true, and 0 for false.

-1 byte thanks to Benoit Esnard!

Brachylog, 4 bytes

ḃo-0

Try it online!

With multiple test cases (😈 is evil and 👼 is not.)

Uses something I discovered recently about the - predicate: its documentation just says "the difference of elements of [input]", but what it actually does is "sum of even-indexed elements (starting from 0th) of input, minus the sum of odd-indexed elements of input".

Here,

ḃ converts the number into an array of binary digits,

o sorts them to bring all the 1s together.

Now, if there were an even number of 1s, there would be an equal number of 1s in even indices and odd indices. So the - after that would give a 0. But if there were an odd number of 1s, there would be an extra 1 sticking out, resulting in the difference being either -1 or 1.

So, finally, we assert that the difference is 0, and get a true or false result according to that. With more flexible output requirements, this could be removed for a 3 byte answer, with 0 as truthy output and -1 and 1 as both falsey outputs.

INTERCAL, 90 65 63 bytes

DOWRITEIN:1
DO:2<-'#0$#65535'~?':1~:1'
DOREADOUT:2
PLEASEGIVEUP

Try it online!

Ungolfed and expanded (for what it's worth) with C style comments.

DO WRITE IN :1 //Store user input in 1
DO :2<-:1~:1 //Select just the ones. So will convert binary 10101 to 111
DO :3<-:?2 //Run the unary xor over the result. Essentially, xor with the right bitshifted
           //(with wraparound) value).
DO :9<-#0$#65535 //Intermingle the 16 bit values of all 0's and all 1's, to create a
                 //32 bit number with 1's in the odd positions.
DO :4<-:9~:3 //It turns out that at this point, evil numbers will have no bits in odd
             //positions, and non-evil numbers will have precisely one bit in an odd
             //position. Therefore, the ~ will return 0 or 1 as appropriate.
PLEASE READ OUT :4 //Politely output
PLEASE GIVE UP //Polite and self explanatory

I had to make a few concessions to make this feasible in INTERCAL. The first is, as with all INTERCAL programs, numerical input must be written out. So if you want to input 707 you would provide SEVEN OH SEVEN.

The second is that INTERCAL doesn't really have proper truthy or falsy value. Instead, it will output the Roman Numeral I (1) if the number is not evil, or a 0 (typically represented as - since Roman Numerals can't normally represent 0).

If you want to flip those so that evil numbers return 1 and non-evil numbers return 0, you can change lines 4 and 5 from the ungolfed version as follows, although it does add 3 bytes.

DO:9<-#65535$#0
DO:4<-#1~:9~3

Attache, 13 12 bytes

Even@Sum@Bin

Try it online!

(Old 13 bytes: Even@1&`~@Bin)

This is a composition of three functions:

1. Bin
2. Sum
3. Even

This checks that the Sum of the Binary expansion of the input is Even.

dc, 18 16 bytes

[2~rd1<M+]dsMx2%

Returns (to the stack) 0 for evil and 1 for not evil

Try it online!

Fairly straightforward - recursively applies the combined quotient/remainder operator ~ to the new quotient and adds all the remainders together, then mods by 2 (after spending two bytes to flip to a standard truthy/falsy).

Edited to reflect consensus that 0 for truthy and 1 for falsy is okay, especially in a language that has no sort of if(boolean) construct.

Python 2, 29 bytes

lambda n:~bin(n).count('1')&1

Try it online!

Returns 1 if True, else 0.

Converts the number to a binary string like '0b11', counts the number of 1s, gets the complement of result, and returns the last bit of the complement (thanks, https://codegolf.stackexchange.com/users/53560/cdlane!) (1 if the original number was even, 0 if it was odd).

x86-16, 3 bytes

NASM listing:

 1                                  parity16:
 2 00000000 30E0                        xor al,ah
 3 00000002 C3                          ret

16-bit integer function arg in AX (which is destroyed), return value in PF.

The hardware calculates the parity of the result for us, in x86's Parity Flag. The caller can use jp / jnp to branch, or whatever they like.

Works exactly like @cschultz's Z80 / 8080 answer; in fact 8086 was designed to make mechanical source-porting from 8080 easy.

Note that PF is only set from the low byte of wider results, so test edi,edi wouldn't work for an x86-64 version. You'd have to horizontal-xor down to 16 bits, or popcnt eax, edi / and al,1 (where 0 is truthy).

C++ (gcc) (-O0),  36  31 bytes

int f(int i){i=!i||i%2-f(i/2);}

Try it online!

C++ (clang), 35 bytes

int f(int i){return!i||i%2-f(i/2);}

Try it online!

Here is my first attempt at code golfing, hope I didn't break any rule I might have missed.

Edit:
- Saved 5 bytes thanks to @Jonathan Frech : replaced != by - and return by i= (the last replacement does not seem to work with clang though)
- Since there seems to be a debate whether I should use gcc -O0 abuse, I thought I could just give both versions

SML, 32 Bytes

fun%0=1| %n=(n+ %(n div 2))mod 2

Explaination:

• % is function name
• takes in input in repl and returns 1 if evil, 0 otherwise
• n is input, returns (n + %(n//2)) % 2

Made by 2 bored Carnegie Mellon Students

Forth (gforth), 53 bytes

: f 1 swap begin 2 /mod -rot xor swap ?dup 0= until ;

Try it online!

Explanation

Takes the xor-sum of the digits of the binary form of the number. (repeatedly divides by 2 and xors the remainder with the "sum" value)

Code Explanation

: f              \ begin a new word definition
  1 swap         \ place 1 on the stack below the input (n)
  begin          \ start an indefinite loop
    2 /mod       \ get the quotient and remainder of dividing n by 2
    -rot         \ move the sum and remainder to the top of the stack
    xor          \ xor the sum and remainder
    swap         \ move the quotient back to the top of the stack
    ?dup         \ duplicate if > 0
    0=           \ get "boolean" indicating if quotient is 0
  until          \ end the loop if it is, otherwise go back to the beginning
;                \ end the word definition

Java 8, 40 36 bytes

n->n.toString(n,2).chars().sum()%2<1

-4 bytes thanks to @Okx for something I shouldn't have forgotten myself..

Try it online.

Explanation:

n->                // Method with Integer parameter and boolean return-type
  n.toString(n,2)  //  Convert the integer to a binary String
   .chars()        //  Convert that to an IntStream of character-encodings
   .sum()          //  Sum everything together
    %2<1           //  And check if it's even

Note that the character encoding for 0 and 1 are 48 and 49, but summing them and taking modulo-2 still holds the correct results because 48%2 = 0 and 49%2 = 1.

Perl 6, 21 bytes

*.base(2).comb(~1)%%2

Test it

Expanded:

*\        # WhateverCode lambda (this is the parameter)
.base(2)  # Str representing the binary
.comb(~1) # find the "1"s

%% 2      # is the count of "1"s divisible by 2?

Retina 0.8.2, 28 bytes

.+
$*
+`(1+)\1
$+0
0

11

^$

Try it online! Link includes test cases. Explanation:

.+
$*

Convert to unary.

+`(1+)\1
$+0

Partial binary conversion (leaves extra zeroes).

0

Delete all the zeros.

11

Modulo the ones by two.

^$

Test whether the result is zero.

x86 Assembly, 12 11 bytes

F3 0F B8 44 24 04  popcnt      eax,dword ptr [esp+4] ; Load EAX with the number of ones in arg
F7 D0              not         eax ; One's complement negation of EAX
24 01              and         al,1 ; Isolate bottom bit of EAX
C3                 ret             

-1 byte thanks to @ceilingcat's suggestion

Bash + GNU utilities, 33

dc -e2o?p|tr -d 0|wc -c|dc -e?2%p

Try it online!

Reads input from STDIN. Outputs 1 for True and 0 for False.

• dc converts input to a binary string
• tr removes zeros
• wc counts remaining ones (and trailing newline, which corrects sense of logic
• dc calculates count mod 2 and outputs the answer

Python 2, 28 27 bytes

f=lambda n:n<1or n&1^f(n/2)

Try it online!

Returns a truthy value if exactly one of the ones-bit is a 1 and the result of calling this function on n/2 is truthy is true (or n==0). It works because n/2 is equivalent to a right bitshift with floor division (so Python 2 only).

Alternate version, also 28 27 bytes

g=lambda n:n<1or g(n&n-1)^1

Try it online!

Based on the K&R method of counting set bits referenced by vazt.

Both of these could be two bytes shorter if the output allowed falsey to mean evil.

Edit: Thanks to Amphibological for saving a byte!

APL (Dyalog Unicode), 10 bytes^SBCS

Anonymous tacit function. Can take any array of integers as argument.

≠⌿1⍪2∘⊥⍣¯1

Try it online!

2∘⊥⍣¯1 convert to binary, using as many digits as needed by the largest number, separate digits along primary axis

1⍪ prepend ones along the primary axis

≠⌿ XOR reduction along the primary axis

J, 9 bytes

Anonymous tacit function. Can take any integer array as argument.

1-2|1#.#:

Try it online!

1- one minus (i.e. logical negation of)

2| the mod-2 of

1#. the sum (lit. the base-1 evaluation) of

#: the binary representation